'''
快的在环中走了慢的的二倍距离
求节点可以第三个指针和慢的同步走就ok了
'''
class Solution:
    def EntryNodeOfLoop(self, pHead):
        # write code here
        fast = pHead
        slow = pHead
        while fast.next:
            fast = fast.next.next
            slow = slow.next
            if fast == slow:
                slowPro = pHead
                while slowPro != slow:
                    slowPro = slowPro.next
                    slow = slow.next
                return slowPro
        return